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 July 20th, 2010, 10:23 PM #1 Newbie   Joined: May 2010 Posts: 10 Thanks: 0 The Lagrange multiplier method 1. Use the Lagrange multiplier method to solve the following. [assume that the method works, i.e. that the critical point you find is the right kind of extremum] (a) min: (x^2) + xy + (y^2) subject to: x+y=10 (b) max: ?x + 2?y subject to: 2x + y = 18 The solutions are: (a) min=75 at (5,5) (b) max=9 at (1,16) But how to get to the solutions? Can someone solve and explain for me please? 2. Use the Lagrange multiplier method to find the closest and furthest points on the unit sphere in R^3 to the point (2; 2; 2). That is, use the Lagrange multiplier method on each of the two problems: max/min : (||(2,2,2)-(x,y,z)||)^2 subject to: (||(x,y,z)||)^2 =1 Note: "||" above are absolute value brackets. If you apply the method correctly, you should get two answers, one of which will be the minimum and one of which will be the maximum. Can someone please solve this for me? At least I have an idea about 1. but this is ticking me off. Thank you! July 21st, 2010, 05:25 AM #2 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: The Lagrange multiplier method It sounds like you need an introduction to the basics of Lagrange multipliers. See http://www.slimy.com/~steuard/teaching/ ... range.html and http://en.wikipedia.org/wiki/Lagrange_multipliers Tags lagrange, method, multiplier Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post MMCS Calculus 3 December 7th, 2013 07:44 AM Brazen Calculus 1 January 15th, 2013 10:29 AM trey01 Applied Math 2 March 25th, 2012 07:14 AM OSearcy4 Calculus 2 October 16th, 2009 01:44 PM roonaldo17 Calculus 0 November 16th, 2008 11:27 AM

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